Verification theorems and feedback controls
Turn a smooth HJB candidate into a certified lower bound and construct an admissible minimizing feedback.
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The question for today
What you’ll understand
- Prove the cost inequality with Itô’s formula and the minimizing HJB sign.
- Identify the integrability and closed-loop conditions needed for equality.
- Use an exact squared-residual identity to quantify a policy’s suboptimality.
Solving a differential equation is not yet the same as solving a control problem. Verification compares a proposed function with the expected cost of every admissible action, then identifies an admissible action that attains equality. Its proof exposes where smoothness, integrability, and feedback existence enter. An affine terminal-cost problem gives an exact performance gap, so we can check the inequality direction and measure the loss from a specific imperfect policy.
Every admissible policy gives an inequality
Suppose v is continuous up to the terminal time, has one time derivative and two spatial derivatives before it, satisfies the minimizing HJB equation, and matches the terminal cost. For each admissible u, the HJB infimum implies . Apply Itô's formula to v along the controlled state and add the accumulated running cost. The resulting process has nonnegative drift, giving a submartingale comparison after the stochastic integral is justified.
Taking expectations and using terminal matching gives , hence a lower bound on the minimizing value. This sign is easy to reverse when switching from reward maximization to cost minimization. The comparison should hold for all controls in the declared class, not only for the proposed optimizer. Otherwise it proves a local comparison or a restricted policy result, rather than a certificate for the full value.
Local martingales and feedback selectors need care
If the stochastic integral has finite expected integrated squared integrand, it is a true martingale and its expectation vanishes directly. More generally, stop the state on increasing bounded regions, apply the identity to the localized process, and justify passage to the horizon using uniform integrability or a suitable domination and growth estimate. A local martingale need not have zero expectation at an unrestricted horizon. Polynomial growth of v therefore needs matching state moments and derivative estimates.
To achieve equality, select an action that attains the HJB infimum measurably and require the resulting closed-loop SDE to possess an admissible solution. Compact actions and continuity can support a measurable selector, but they do not automatically make a discontinuous selected feedback globally Lipschitz. One may need a separate existence theorem or only epsilon-optimal controls. Once an admissible selector exists and the martingale conditions hold, its HJB drift vanishes and the bound becomes equality.
A completed square can certify the whole policy class
Consider drift controls in the interval from minus one to one, unit additive diffusion, terminal cost equal to the terminal state, and running effort . The candidate is affine in state. Its HJB minimizer is the constant minus one. The diffusion term vanishes because the second spatial derivative is zero, while the stochastic integral is simply a Brownian increment and is square integrable on the finite horizon.
The nonnegative drift becomes half the square of u plus one. Integrating yields an exact difference between any admissible policy cost and the candidate. Bounded controls give the needed state moments even though the linear terminal cost can be negative. The square identifies the optimizer up to product-time-probability null sets and also distinguishes genuine optimality from a policy whose residual merely looks small at one state or one instant.
Work it through
Verify an affine candidate and compare three policies
Take , initial state three, dynamics , and controls in . Minimize . Verify the candidate and calculate the costs of the constant controls minus one, zero, and one.
The candidate is , with time derivative one half, state derivative one, and second derivative zero. Its controlled HJB residual is .
The minimum residual is zero at u equal to minus one, which is admissible. The Brownian stochastic integral has finite second moment on this horizon, so taking expectations in the verification identity is valid without an unproved local-martingale step.
The initial candidate is two. The constant zero has gap one and therefore cost three. The constant one has gap four and therefore cost six. Directly calculating terminal means and effort costs gives the same three values.
The optimal value is two, achieved by the constant control −1. The other two policies cost three and six, exactly as the residual-square certificate predicts.
The living notebook
Explore the model
Interactive lab · optimization
Reduce a verified residual rather than assume optimality
For the residual z=u+1, the running gap is z²/2. Pause and Reset, then set step size to 0.5 and initial position to 1. One Step gives z=0.5, corresponding to u=−0.5 and gap rate 0.125. For a fresh comparison, Pause and Reset, then set step size to 1 and initial position to 1; one Step reaches z=0. This is gradient descent on one quadratic residual, not a diffusion or a proof of stochastic verification. Parameter edits retain elapsed time, so reset both settings before each comparison.
- Position x0
- 2
- Objective x²/2
- 2
- Iteration behavior
- Stable step
Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA residual calculation measures suboptimality only after the verification identity and admissibility have been established.
Pause & reason
Check your understanding
Your turn
Try it yourself
In the affine problem, start at state minus one with one unit of time remaining and use constant control −0.5. Compute its cost, the optimal value, and their gap.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The terminal mean is initial state plus control times duration. Add half the integrated squared action and compare with the residual-square identity.
Compare with the solution
The terminal mean is −1.5 and the effort contribution is 0.125, so the policy cost is −1.375. The optimal value is −1−0.5=−1.5. Their gap is 0.125, equal to one half times the one-unit integral of (−0.5+1)². Negative values are allowed because the terminal cost is linear, not a nonnegative state penalty.
Keep the idea
Take it with you
- A minimizing HJB candidate gives a lower bound once the stochastic expectation steps are valid.
- Feedback selection, closed-loop existence, and integrability are separate verification requirements.
- An exact residual identity can quantify a policy’s loss over the entire horizon.
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